Answer
Verified
387.9k+ views
Hint: We’ll first write the conditional equation for a, b, c, and then will note the value of b and c in terms of a and common ratio to create a relation in a, b, c and then will find the required answer with help of those equations.
Complete step by step answer:
Given data: a, b, c are in G.P.
From the given data i.e. a, b, c are in G.P., we can say that the common ratio will remain the same
$
\dfrac{{\text{b}}}{{\text{a}}}{\text{ = }}\dfrac{{\text{c}}}{{\text{b}}}..............{\text{(i)}} \\
\Rightarrow {{\text{b}}^{\text{2}}}{\text{ = ac}}.........{\text{(ii)}} \\
$
Let the common ratio be ‘r’
$\therefore {\text{b = ar}}$
Squaring both sides of the above equation
$
{{\text{b}}^{\text{2}}}{\text{ = }}{\left( {{\text{ar}}} \right)^{\text{2}}} \\
\Rightarrow {{\text{b}}^{\text{2}}}{\text{ = }}{{\text{a}}^{\text{2}}}{{\text{r}}^{\text{2}}} \\
$
Also,${\text{c = a}}{{\text{r}}^{\text{2}}}$
Squaring both sides of the above equation
$
{{\text{c}}^{\text{2}}}{\text{ = }}{\left( {{\text{a}}{{\text{r}}^{\text{2}}}} \right)^{\text{2}}} \\
\Rightarrow {{\text{c}}^{\text{2}}}{\text{ = }}{{\text{a}}^{\text{2}}}{{\text{r}}^{\text{4}}} \\
$
From the value of ${{\text{a}}^{\text{2}}}{\text{,}}{{\text{b}}^{\text{2}}}{\text{,}}{{\text{c}}^{\text{2}}}$ we can see that they are in G.P. with a common ratio of ${\text{'}}{{\text{r}}^{\text{2}}}{\text{'}}$
Common ratio=$\dfrac{{{{\text{b}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}}{\text{ = }}\dfrac{{{{\text{c}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}}{\text{ = }}{{\text{r}}^{\text{2}}}$
Therefore, option(A) is correct
Note: We can also verify our solution with the help of an example let say 2,4,8 where
a=2
b=4
c=8
for option(A) ${{\text{a}}^{\text{2}}}{\text{,}}{{\text{b}}^{\text{2}}}{\text{,}}{{\text{c}}^{\text{2}}}$=${\text{4,16,64}}$
since $\dfrac{{{\text{16}}}}{{\text{4}}}{\text{ = }}\dfrac{{{\text{64}}}}{{{\text{16}}}}{\text{ = 4}}$
therefore, ${{\text{a}}^{\text{2}}}{\text{,}}{{\text{b}}^{\text{2}}}{\text{,}}{{\text{c}}^{\text{2}}}$are in G.P.
(A)${{\text{a}}^{\text{2}}}{\text{,}}{{\text{b}}^{\text{2}}}{\text{,}}{{\text{c}}^{\text{2}}}$ are in G.P.
Therefore option (A) is correct
for option(B) ${{\text{a}}^{\text{2}}}{\text{(b + c),}}{{\text{c}}^{\text{2}}}{\text{(a + b),}}{{\text{b}}^{\text{2}}}{\text{(a + c)}}$=${{\text{2}}^{\text{2}}}{\text{(4 + 8),}}{{\text{8}}^{\text{2}}}{\text{(2 + 4),}}{{\text{4}}^{\text{2}}}{\text{(2 + 8)}}$
${\text{48,386,160}}$
since $\dfrac{{{\text{386}}}}{{{\text{48}}}}{\text{ = }}\dfrac{{{\text{193}}}}{{{\text{24}}}} \ne \dfrac{{{\text{160}}}}{{{\text{386}}}}{\text{ = }}\dfrac{{{\text{80}}}}{{{\text{193}}}}$
therefore, ${{\text{a}}^{\text{2}}}{\text{(b + c),}}{{\text{c}}^{\text{2}}}{\text{(a + b),}}{{\text{b}}^{\text{2}}}{\text{(a + c)}}$are not in G.P.
for option(C) $\dfrac{{\text{a}}}{{{\text{b + c}}}}{\text{,}}\dfrac{{\text{b}}}{{{\text{c + a}}}}{\text{,}}\dfrac{{\text{c}}}{{{\text{a + b}}}}$=$\dfrac{{\text{2}}}{{{\text{4 + 8}}}}{\text{,}}\dfrac{{\text{4}}}{{{\text{8 + 2}}}}{\text{,}}\dfrac{{\text{8}}}{{{\text{2 + 4}}}}$
$\dfrac{{\text{1}}}{{\text{6}}}{\text{,}}\dfrac{{\text{2}}}{{\text{5}}}{\text{,}}\dfrac{{\text{4}}}{{\text{3}}}$
since $\dfrac{{\dfrac{{\text{2}}}{{\text{5}}}}}{{\dfrac{{\text{1}}}{{\text{6}}}}}{\text{ = }}\dfrac{{{\text{12}}}}{{\text{5}}} \ne \dfrac{{\dfrac{{\text{4}}}{{\text{3}}}}}{{\dfrac{{\text{2}}}{{\text{5}}}}}{\text{ = }}\dfrac{{{\text{10}}}}{{\text{3}}}$
therefore, $\dfrac{{\text{a}}}{{{\text{b + c}}}}{\text{,}}\dfrac{{\text{b}}}{{{\text{c + a}}}}{\text{,}}\dfrac{{\text{c}}}{{{\text{a + b}}}}$are not in G.P.
Complete step by step answer:
Given data: a, b, c are in G.P.
From the given data i.e. a, b, c are in G.P., we can say that the common ratio will remain the same
$
\dfrac{{\text{b}}}{{\text{a}}}{\text{ = }}\dfrac{{\text{c}}}{{\text{b}}}..............{\text{(i)}} \\
\Rightarrow {{\text{b}}^{\text{2}}}{\text{ = ac}}.........{\text{(ii)}} \\
$
Let the common ratio be ‘r’
$\therefore {\text{b = ar}}$
Squaring both sides of the above equation
$
{{\text{b}}^{\text{2}}}{\text{ = }}{\left( {{\text{ar}}} \right)^{\text{2}}} \\
\Rightarrow {{\text{b}}^{\text{2}}}{\text{ = }}{{\text{a}}^{\text{2}}}{{\text{r}}^{\text{2}}} \\
$
Also,${\text{c = a}}{{\text{r}}^{\text{2}}}$
Squaring both sides of the above equation
$
{{\text{c}}^{\text{2}}}{\text{ = }}{\left( {{\text{a}}{{\text{r}}^{\text{2}}}} \right)^{\text{2}}} \\
\Rightarrow {{\text{c}}^{\text{2}}}{\text{ = }}{{\text{a}}^{\text{2}}}{{\text{r}}^{\text{4}}} \\
$
From the value of ${{\text{a}}^{\text{2}}}{\text{,}}{{\text{b}}^{\text{2}}}{\text{,}}{{\text{c}}^{\text{2}}}$ we can see that they are in G.P. with a common ratio of ${\text{'}}{{\text{r}}^{\text{2}}}{\text{'}}$
Common ratio=$\dfrac{{{{\text{b}}^{\text{2}}}}}{{{{\text{a}}^{\text{2}}}}}{\text{ = }}\dfrac{{{{\text{c}}^{\text{2}}}}}{{{{\text{b}}^{\text{2}}}}}{\text{ = }}{{\text{r}}^{\text{2}}}$
Therefore, option(A) is correct
Note: We can also verify our solution with the help of an example let say 2,4,8 where
a=2
b=4
c=8
for option(A) ${{\text{a}}^{\text{2}}}{\text{,}}{{\text{b}}^{\text{2}}}{\text{,}}{{\text{c}}^{\text{2}}}$=${\text{4,16,64}}$
since $\dfrac{{{\text{16}}}}{{\text{4}}}{\text{ = }}\dfrac{{{\text{64}}}}{{{\text{16}}}}{\text{ = 4}}$
therefore, ${{\text{a}}^{\text{2}}}{\text{,}}{{\text{b}}^{\text{2}}}{\text{,}}{{\text{c}}^{\text{2}}}$are in G.P.
(A)${{\text{a}}^{\text{2}}}{\text{,}}{{\text{b}}^{\text{2}}}{\text{,}}{{\text{c}}^{\text{2}}}$ are in G.P.
Therefore option (A) is correct
for option(B) ${{\text{a}}^{\text{2}}}{\text{(b + c),}}{{\text{c}}^{\text{2}}}{\text{(a + b),}}{{\text{b}}^{\text{2}}}{\text{(a + c)}}$=${{\text{2}}^{\text{2}}}{\text{(4 + 8),}}{{\text{8}}^{\text{2}}}{\text{(2 + 4),}}{{\text{4}}^{\text{2}}}{\text{(2 + 8)}}$
${\text{48,386,160}}$
since $\dfrac{{{\text{386}}}}{{{\text{48}}}}{\text{ = }}\dfrac{{{\text{193}}}}{{{\text{24}}}} \ne \dfrac{{{\text{160}}}}{{{\text{386}}}}{\text{ = }}\dfrac{{{\text{80}}}}{{{\text{193}}}}$
therefore, ${{\text{a}}^{\text{2}}}{\text{(b + c),}}{{\text{c}}^{\text{2}}}{\text{(a + b),}}{{\text{b}}^{\text{2}}}{\text{(a + c)}}$are not in G.P.
for option(C) $\dfrac{{\text{a}}}{{{\text{b + c}}}}{\text{,}}\dfrac{{\text{b}}}{{{\text{c + a}}}}{\text{,}}\dfrac{{\text{c}}}{{{\text{a + b}}}}$=$\dfrac{{\text{2}}}{{{\text{4 + 8}}}}{\text{,}}\dfrac{{\text{4}}}{{{\text{8 + 2}}}}{\text{,}}\dfrac{{\text{8}}}{{{\text{2 + 4}}}}$
$\dfrac{{\text{1}}}{{\text{6}}}{\text{,}}\dfrac{{\text{2}}}{{\text{5}}}{\text{,}}\dfrac{{\text{4}}}{{\text{3}}}$
since $\dfrac{{\dfrac{{\text{2}}}{{\text{5}}}}}{{\dfrac{{\text{1}}}{{\text{6}}}}}{\text{ = }}\dfrac{{{\text{12}}}}{{\text{5}}} \ne \dfrac{{\dfrac{{\text{4}}}{{\text{3}}}}}{{\dfrac{{\text{2}}}{{\text{5}}}}}{\text{ = }}\dfrac{{{\text{10}}}}{{\text{3}}}$
therefore, $\dfrac{{\text{a}}}{{{\text{b + c}}}}{\text{,}}\dfrac{{\text{b}}}{{{\text{c + a}}}}{\text{,}}\dfrac{{\text{c}}}{{{\text{a + b}}}}$are not in G.P.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE